Mixing
Why do I get an error when I put a call for the output of a function (defined within a function) *after* the...
My code does what I want it to do, but I want to understand why it does what it does. I find that in MATLAB if I call for the output of a function after the definition of the function, I get an error, but not if I call it before the function's definition. My impression is that MATLAB is putting the cart before the horse, and I'd like to know why I am mistaken in that impression.
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
return
outt=f_inside(arg_in);
return
The expression out=f_testFunctionWithinFunction(5) yields -0.9589=sin(5). However, [out,outt]=f_testFunctionWithinFunction(5) yields an error message, due to asking for "outt". Why?
matlab function nested
edited Jan 2 at 6:35
rahnema1
10.6k2923
asked Jan 2 at 3:57
ignoramusignoramus
6
add a comment |
My code does what I want it to do, but I want to understand why it does what it does. I find that in MATLAB if I call for the output of a function after the definition of the function, I get an error, but not if I call it before the function's definition. My impression is that MATLAB is putting the cart before the horse, and I'd like to know why I am mistaken in that impression.
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
return
outt=f_inside(arg_in);
return
The expression out=f_testFunctionWithinFunction(5) yields -0.9589=sin(5). However, [out,outt]=f_testFunctionWithinFunction(5) yields an error message, due to asking for "outt". Why?
matlab function nested
edited Jan 2 at 6:35
rahnema1
10.6k2923
asked Jan 2 at 3:57
ignoramusignoramus
6
If you want to know how to define functions you can read about function. Alsolocal functionsshould be distinguished fromnested functions.
– rahnema1
Jan 2 at 7:38
add a comment |
My code does what I want it to do, but I want to understand why it does what it does. I find that in MATLAB if I call for the output of a function after the definition of the function, I get an error, but not if I call it before the function's definition. My impression is that MATLAB is putting the cart before the horse, and I'd like to know why I am mistaken in that impression.
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
return
outt=f_inside(arg_in);
return
The expression out=f_testFunctionWithinFunction(5) yields -0.9589=sin(5). However, [out,outt]=f_testFunctionWithinFunction(5) yields an error message, due to asking for "outt". Why?
matlab function nested
edited Jan 2 at 6:35
rahnema1
10.6k2923
asked Jan 2 at 3:57
ignoramusignoramus
6
My code does what I want it to do, but I want to understand why it does what it does. I find that in MATLAB if I call for the output of a function after the definition of the function, I get an error, but not if I call it before the function's definition. My impression is that MATLAB is putting the cart before the horse, and I'd like to know why I am mistaken in that impression.
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
return
outt=f_inside(arg_in);
return
The expression out=f_testFunctionWithinFunction(5) yields -0.9589=sin(5). However, [out,outt]=f_testFunctionWithinFunction(5) yields an error message, due to asking for "outt". Why?
matlab function nested
matlab function nested
edited Jan 2 at 6:35
rahnema1
10.6k2923
asked Jan 2 at 3:57
ignoramusignoramus
6
edited Jan 2 at 6:35
rahnema1
10.6k2923
asked Jan 2 at 3:57
ignoramusignoramus
6
edited Jan 2 at 6:35
rahnema1
10.6k2923
edited Jan 2 at 6:35
rahnema1
10.6k2923
edited Jan 2 at 6:35
rahnema1
10.6k2923
10.6k2923
asked Jan 2 at 3:57
ignoramusignoramus
6
asked Jan 2 at 3:57
ignoramusignoramus
6
asked Jan 2 at 3:57
ignoramusignoramus
6
6
If you want to know how to define functions you can read about function. Alsolocal functionsshould be distinguished fromnested functions.
– rahnema1
Jan 2 at 7:38
add a comment |
If you want to know how to define functions you can read about function. Alsolocal functionsshould be distinguished fromnested functions.
– rahnema1
Jan 2 at 7:38
If you want to know how to define functions you can read about function. Also
local functions should be distinguished from nested functions.– rahnema1
Jan 2 at 7:38
If you want to know how to define functions you can read about function. Also
local functions should be distinguished from nested functions.– rahnema1
Jan 2 at 7:38
add a comment |
1 Answer
1
active
oldest
votes
you should end functions with end, instead of return. try:
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
disp('test')
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
end
outt=f_inside(arg_in);
end
answered Jan 2 at 7:29
Yuval HarpazYuval Harpaz
735512
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54001039%2fwhy-do-i-get-an-error-when-i-put-a-call-for-the-output-of-a-function-defined-wi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
you should end functions with end, instead of return. try:
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
disp('test')
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
end
outt=f_inside(arg_in);
end
answered Jan 2 at 7:29
Yuval HarpazYuval Harpaz
735512
add a comment |
you should end functions with end, instead of return. try:
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
disp('test')
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
end
outt=f_inside(arg_in);
end
answered Jan 2 at 7:29
Yuval HarpazYuval Harpaz
735512
add a comment |
you should end functions with end, instead of return. try:
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
disp('test')
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
end
outt=f_inside(arg_in);
end
answered Jan 2 at 7:29
Yuval HarpazYuval Harpaz
735512
you should end functions with end, instead of return. try:
function [out,outt]=f_testFunctionWithinFunction(arg_in)
% function has to be invoked BEFORE its definition.
disp('test')
out=f_inside(arg_in);
function use=f_inside(argg_in),
use=sin(argg_in);
end
outt=f_inside(arg_in);
end
answered Jan 2 at 7:29
Yuval HarpazYuval Harpaz
735512
answered Jan 2 at 7:29
Yuval HarpazYuval Harpaz
735512
answered Jan 2 at 7:29
Yuval HarpazYuval Harpaz
735512
answered Jan 2 at 7:29
Yuval HarpazYuval Harpaz
735512
735512
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54001039%2fwhy-do-i-get-an-error-when-i-put-a-call-for-the-output-of-a-function-defined-wi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
If you want to know how to define functions you can read about function. Also
local functionsshould be distinguished fromnested functions.– rahnema1
Jan 2 at 7:38